Answer
$x=\left\{ 2-\sqrt{6},2+\sqrt{6} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
(2x+1)(x-4)=x
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula to solve for $x.$
$\bf{\text{Solution Details:}}$
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
2x(x)+2x(-4)+1(x)+1(-4)=x
\\\\
2x^2-8x+x-4=x
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
2x^2+(-8x+x-x)-4=0
\\\\
2x^2-8x-4=0
.\end{array}
In the equation above, $a=
2
,$ $b=
-8
,$ and $c=
-4
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(2)(-4)}}{2(2)}
\\\\
x=\dfrac{8\pm\sqrt{64+32}}{4}
\\\\
x=\dfrac{8\pm\sqrt{96}}{4}
\\\\
x=\dfrac{8\pm\sqrt{16\cdot6}}{4}
\\\\
x=\dfrac{8\pm\sqrt{(4)^2\cdot6}}{4}
\\\\
x=\dfrac{8\pm4\sqrt{6}}{4}
\\\\
x=\dfrac{\cancel4^2\pm\cancel4^1\sqrt{6}}{\cancel4^1}
\\\\
x=2\pm\sqrt{6}
.\end{array}
The solutions are $
x=\left\{ 2-\sqrt{6},2+\sqrt{6} \right\}
.$
